ssc mains (maths) mock test-2 (solution)
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8/10/2019 Ssc Mains (Maths) Mock Test-2 (Solution)
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MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR
DILSHAD GARDEN ROHINI
BADARPUR BORDER
SSC (Tier-II) - 2013 (Mock Test Paper - 2) [SOLUTION]
1. (C)4847
4847
484747 4747
R 48
47
48
47 =
48
94 R = 46
2. (D)Let the first odd integer = x
So, the second odd integer = x+ 2Now,
Difference between the squares of these twoconsecutive odd integers
= (x + 2)2 x2
= x2+ 4 + 4x x2
= 4( + 1)x
will be divisibleby 2 as ( + 1) will be evenx
4(x+ 1) will be divisible by
4 2 i.e. by 83. (D)Let x= the larger number
and y= the smaller numberNow,
ATQ, x y= 2395 ... (i)
andy
x; Quotient = 7 and Remainder = 25
x= 7y+ 25 ... (ii)Now on putting value of y from (i) in (ii)
We get,x= 7(x 2395) + 25
or, x= 7x 16765 + 25or, 6x= 16765 25
x=6
16740= 2790
4. (C) Let A, B, C and D are four consecutive prime
numbers in descending order ( A > B > C > D)So, ATQ,
A B C = 2431 ... (i)
and B C D = 1001 ... (ii)Now, both 2431 and 1001 are divisible by 11.
i.e. 11 is a common factor of both 2431 and1001.
andB & C are the common factors of 2431and 1001.
Out of B & C, one must be 11.
So, from (i),
Product of other two prime nos. =11
2431
= 221and 221 = 17 13
Out of A, B and C, one is 17, one is 13and one is 11.
and A > B > C
A = 17, B = 13 and C = 11
So, the largest given prime no. = 17
5. (A) 182221164 3333
= 3 3 3( 2 2) (2 2 1) (1 1)
= 2323 )1()122()2(
= 23 )12(
So, square root of )1164( 33
i.e )1164( 33 = 123
6. (B) 0.5 =2
1
10
5
2.1 =10
21
2.8 =10
28=
5
14
So, LCM of 0.5, 2.1 & 2.8
= LCM of10
21,
2
1&
5
14
=5&10,2ofHCF
14&21,1ofLCM=
1
42= 42
7. (C) Required multiplication
= 2 4 6 8 10 12 14 16 18 20 5 10 15 20 25
Here, nos. (o) = no. of complete pairs of 2 & 5.= no. of 5 [here n(5) < n(2)]
= 7
8. (C)42.03.03.07.07.0
37.03.03.03.03.07.07.07.0
=0.7 0.7 0.7 0.3 0.3 0.3 3 0.7 0.3 1
0.7 0.7 0.3 0.3 2 0.7 0.3
=
3 2
2 2
(0.7) (0.3) 0.3 0.7 0.3 (0.7 0.3)
(0.7) (0.3) 2 0.7 0.3
= 11
1
1
1
)3.07.0(
)3.07.0(2
3
2
3
9. (D)each three distance is same.Let, S
1, S
2and S
3respectively are the speeds
with which three distances have beencovered.
So,
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Average Speed =S S S
S S S S S S 1 2 3
1 2 2 3 3 1
3
=154040303015
4030153
=6001200450
4030153
=54000
2250= 24 km/hr
10. (B) A + C =37
22part
B =37
221 part =
37
15part
and,
B + C = 37
21
part
or, C15 21
27 37 C =
37
6
37
15
37
21 part
So,
Wage of C = 925037
6 = Rs. 1500
11. (C) Let,
(A + B) xdays
So, A alone (x+ 8) days
and, B alone (x+ 18) daysIn such type of questions
The no. of days taken by A and B working
together to do the work = 188 days
= 144 days
= 12 days
12. (*)Ratio of speedRatio of durationRatio of distancecovered during
any same durationof time
== := :
A
236
B
111
( Distance = Speed time )
13. (A) 600 km
570 km 30 km
B' BA
Distance covered by B before movement of A.i.e. Distance covered by B in 20 minutes
=
60
2090 km
= 30 km
When train from station A starts tomove; the another train will be B' anddistance between A & B'
= (600 30) km = 570 km
Now,
Relative speeds of trains= (100 + 90) km/hr
= 190 km /hr
So, Time taken by each train to reach
each other =
190
570hr. = 3 hrs.
And in 3 hours, distance travelled by A
= (100 3) km = 300 kms
Both train will cross each other at adistance 300 km & from A i.e. at the exactmiddle point of A and B.
14. (B) Let x= the required speed
Home Office
54 km/hr hr.
xkm/hr hr.
1260
12 +660
60
1254 =
60
18x
x= 36 km/hr
15. (C)
Ratio of speedsTime taken
As the distance is same.
(where x time taken by B to coverthe distance)
:
A
3( + 8)hrx
B
5x hr.
3(x+ 8) = 5x
3x+ 24 = 5x
x = 12 hr.
So,
Actual time taken by A to cover thedistance i.e. (x+ 8) hr. = (12 + 8) hrs
= 20 hrs.
16. (C) Sum of wrongly entered marks
= 73 + 78 + 80
= 231and Sum of correct marks
= 63 + 70 + 82
= 215
i.e. Sum of corrected marks
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The average of above 5 consecutive integers
= m=2
4 nn m = n+ 2
Now,
n + 2 n + 3 n + 4 n + 5 n + 6 n + 7
The average of above 6 consecutive integers
=2
72 nn=
2
92 n
=2
542 n=
2
5)2(2 n=
2
52 m
18. (D) Total money spent by all of them
= (18 25) + {1 (25 + 15)}
= 450 + 40
= 490
19. (C) Let x= no. of girls
So,600 11 yr 9 months
= x 11 yrs + (600 x) 12 yrs
7050 yr. = (11x + 7200 12x)yr.
x = 7200 7050
= 150
20. (A) Vessel '1' Vessel '2'
Water WaterMilk Milk3 5: :2 3
Ratio of contents of vessel '1' to vessel '2' tomix with each other = 1 : 2
Ratio of water and milk in the resulting
mixture =
28
3
5
2
28
5
5
3
=
40
301640
5024
=46
74
21. (B) Initial Ratio = 4 : 6 : 9
Initial nos. = 4x, 6x, 9x
Final Ratio = 7 : 9 : 12
(on increasing 12 students in each class)
126
124
x
x
= 9
7
or, 36x+ 108 = 42x+ 84
6x = 24 x= 4
Total no . of students in the threeclasses before the increase = 4x+ 6x+ 9x
= 19x= 76
22. (B) ATQ
10% of 1st no. = 6
or,10
1of 1stno. 6 1stno. = 60
and, Ratio of 1stno. to 2ndno. = 4 : 3
Nos. are 60 & 45
Now, 20% of 2ndno. = 20% of 45 = 9
& 5% of 1stno. = 5% of 60 = 3
And, 9 is 3 times of 3.
20% of 2ndno. is 3 times of 5% of thefirst no.
23. (D) 1st
no. : 2nd
no. &2nd
no. : 3rd
no.= 3 : 2 = 3 : 2
1stno. : 2ndno. : 3rdno.
= 9 : 6 : 4
So, Let the nos. are 9x, 6x& 4x
ATQ, (9x)2+ (6x)2+ (4x)2+ 532
133x2= 532
x= 2
So, the 2ndno.
i.e. 6x= 6 2 = 12
24. (?) Table Chair
20% profit 25% profit
1 1
10
2 2
5
3 3
3
2
3 3
3
1
3 3
profit1
323
1 23 3
1300
= Rs. 433.33
1300
2 : :
:
:
1
1: 2
1 1
Cost price of table = Rs. 433.33
Alternative method:-
Let,
x = C.P. of table
(1300 x) = C.P. of chair
Now, ATQ,
Profit on table + profit on chair
= Total profit
or, 20% of x+ 25% of (1300 x)
= %3
123 of 1300
or, )1300(100
25
100
20xx = 1300
1003
70
or,3
910
44
1300
5
xx
or, 3253
910
45
xx
or,20
54 xx=
3
975910
or,20
x =
3
65
x= 3
2065
= Rs. 433.33
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25. (C) A's share = Rs. 12000
=8
3of (100% 20%) of total
profit
=83 of 80% of total profit
80% of total profit = Rs. 12000 3
8
So, Total profit (i.e. 100%)
= Rs. 12000 3
8
100
80
= Rs. 40000
26. (A) Let the third no. be 100.
So, A and B will be 80 and 72 respectively.
So, required % =Aofvalue
B&Aofdifference 100%
= %10080
8 = 10%
27. (B) Total no. of illiterate persons in the town
= no. of literate men + no. of illiteratewomen
=
31125083
40of%)24%100( +
31125083
43
of%)8%100(
= 31125083
43
100
92311250
83
40
100
76
= 114000 + 148350
= 262350
28. (B) Let x= maximum marks
So, ATQ,
30% of x+ 50 = 320 + 30
or, x100
30= 300
x =30
100300 = 1000 marks
29. (D) Let x= man's initial income
So, Man's increased income = (x+ 1200)
Now, ATQ,
12% of x = 10% of (x+ 1200)
12x = 10(x + 1200)
or, 12x 10x = 12000
x = 6000
So, increased income;
i.e. x+ 1200 = Rs. 6000 + Rs. 1200
= Rs. 7200
30. (D) 20% profit on one and 20% loss on other
net loss
and loss % =100
)20( 2= 4%
Total SP = (100 4)% of total CP 2 4.5 lakh = 96% of total CP
So,
loss i.e. 4% of total CP
= 496
lakh5.42
= Rs. 37500
loss of Rs. 37500
31. (A) Required single discount
= (1 0.9 0.8 0.75) 100%
= (1 0.54) 100%
= 0.46% 100%= 46%
32. (D) SP per mango = Re.10
1
= 140% of CP per mango
So, CP per mango (i.e. 100%)
= Re.140
100
10
1 = Re.
14
1
14 mangoes for Re. 1.
33. (A) Let x= CP of 2ndwatch
So, (1120 x) = CP of 1stwatch
ATQ, Profit = Loss
15% of (1120 x) = 10% of x
15(1120 x) = 10x
25x = 15 1120
x =25
112015 = Rs. 672
34. (D)1
5
2
2
1710% (15 + 2.5)% %
=
15%1 : 2
2 : 1
=
gap = 5% gap = % 2.5%
50 pens (150 50) pens = 100 pens
35. (D) C.P. of 40 article = SP of 25 articles
In such type of questions,
Gain percent = %10025
2540
= %10025
15
= 60%
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36. (B)C.P. S.P.A
B
C
B
C
Rs. 600
=Rs.
= Rs.
Rs. x
Rs. x
Rs. x
Rs. +x
Rs.
Rs.
Now,
Now,
x
120
36
6
6
1
36
36
6
36
x
x
x
x
x
5
100
25
5
5
6
25
25
5
25
= Rs. x25
36
6
5Rs.
5
6x
Rs. x5
6= Rs. 600
So, CP of A i.e. Rs. x = Rs.6
5600
= Rs. 500
37. (*)M.P. Price after 10% discount
10% discount
6% discount
(additional)
Rs.x
Again,
Rs.
Rs.Rs.94
9
9 9x= Rs. 846
x
x100
10
10 10
The original marked price i.e. x
= Rs.994
10100846
= Rs. 1000
38. (B) Single equivalent discount of 20% &
then %4
16 .
= %100
4
2520
%4
25%20
= 25% discount
C.P.
Rs. 100(let)
S.P.
Rs. 120=75% of M.P.
M.P.20% gain 25% discount
So, 100% of MP = 10075
120
= Rs. 160
MP should be above CP by
= %100100
100160
= 60%
39. (C) 20% discount on L.P. 25% discount on
L.P. = Rs. 500
80% value of L.P. 75% value of L.P.
= Rs. 500
5% value of L.P. = Rs. 500
L.P. (i.e. 100%) = Rs.5
100500
= Rs. 10000
So, Tarun bought the TV at 80% of L.P.
= Rs. 10000100
80
= Rs. 8000
40. (B) SI @ 3% p.a. for 4 yrs. = 12% of sum
SI @ 2% p.a. for 5 yrs. = 10% of sum
ATQ,
(12% 10%) of sum = Rs. 150
2% of sum = Rs. 150
sum = Rs.2
100150
= Rs. 7500
41. (C)
For 1st year
For 2nd year
S.I.
Rs. 135Rs. 135
C.I.
Rs. 135
Now,
if r= rate of interest per annum
r% of 135 = 27
r=135
10027 = 20
Also,
20% of the sum = 135
sum =20
100135 = Rs. 675
42. (C)
For 4 years
For 5 years
C.I.Rs. 3840Rs. 3936 = Rs. 3840 + Rs. 96
Now,
if r = rate of interest per annum
r% of 3840 = 96
r = 3840
10096
= 2.5%
43. (B) ATQ,
where P Principal
& t required no. of years
P +20
t
1 > 2P100
or, 250
11
t
or 25
6
t
Now,
25
61
, 2
5
62
, 2
5
63
but 2
5
64
Required least no. of complete yrs. = 4yrs.
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44. (D) 12M + 16B 10 days ... (i)
18M + 50B 5 days
9M + 25B 10 days ... (ii)From (i) & (ii)
We have
12M + 16B = 9M + 25B 3M = 9B
1M = 3B
So,
Efficiency of a man to that of a boy = 3 : 1
45. (A)
1
1
10
12
Prem 10 days
12 days
part of work
part of work
in 1 day
in 1 day
Prem
Also,
Raj
Raj
Now,
Work done by Prem in initial 6 days
=
6
10
1
=5
3part of work
Remaining part of the work
=5
31
=5
2part of the work
Now,
Amount of work done by Prem & Raj
together in 1 day
=
12
1
10
1
=60
11part of the work
So, No. of days taken by Prem & Raj togetherto do the remaining part of the work
=11
22
115
602
60
115
2
days
Raj actually did the work for11
22 days.
46. (*) A B
6 hours 2 hours
Let the capacity of tank = 12 litres
Rate of filling of tank by pipe A
=
6
12litre per hour
= 2 litre per hour
and rate of filling of tank by pipe B
=
2
12litre per hour
= 6 litre per hour
Rate of filling of tank by pipe A & B together= 8 litre per hour
Now,
Pipe B was opened2
1hour earlier than
pipe A.
Volume of tank filled by pipe B in2
1hour
= 3 litre
Remaining part of the tank which isstill empty = (12 3) litre
= 9 litre
Time required to fill 9 litre of tank by
pipe A & B together
=8
9hours
=8
11 hours
= 1 hours 7.5 minutes Required point of time
= 2 : 30 pm + 1 hour 7.5 mins
= 3 : 37.5 pm or 3 : 372
1pm
Alternative method
Part of tank filled by pipe B alone in2
1hour
=2
1
2
1=
4
1part
Time taken by pipe A & B together to fill
the remaining part
=8
9
6
44
3
2
1
6
14
11
hours
Required point of time
= 2 : 30 pm + 1 hour 7.5 min
= 3 : 372
1pm
47. (B) Let xhour = time taken by pipe A alone to empty the pool
2xhr. = time taken by pipe B
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alone to empty the poolSo, Time taken by pipe A & B together to
empty the pool
=xx
xx
2
2
hours
=x
x
3
2 2hours = x
32
hours
Time taken by pipe C alone to empty
the pool =
2
3
2x hours
= x3
4hours
Part of the pool which will be emptywhen A, B & C work together
=x x x
1 1 3
2 4
part
=
x4
324part
=x4
9part
Total time taken by A, B & C workingtogether to empty the pool
=9
4x= 400 minutes
[6 hour 40 minutes = 400 min]
x =4
9400 minutes
= 900 minutes = 15 hours48. (A) 12 km up + 18 km down in 3 hrs.
36 km up + 54 km down in 9 hrs...... (i)
Also, 36 km up + 24 km down in2
16 hrs.
..... (ii)From (i) & (ii), we get,
30 km down in2
12 hours
Sdown
(i.e. SB+ S
C) = 12 km/hour
Also, S
up
(i.e. SB
+ SC
) = 8 km/hourSo,
Speed of current
i.e. SC
=2
SS updown
=2
812 km/hour
= 2 km/hour49. (C) 24, 20, 16, ..........
Let, n= required no. of terms
Now,
Sn = 2
n
{2a+ (n 1)d}
i.e.180 =2
n{2 (24) + (n1)4}
or, 180 =2
n(48 + 4n 4)
or, 360 = 4n2
52nor, 4n2 52n 360 =0
n= 18
50. (D) 3 18 12 72 66 396 ?3 6 18 6 12 6 72 6 66 6 396 6
The missing no. = 396 6
= 390
51. (A)
21
xx = 3
x
x1
= 3
On cubing both the sides we get,
31
xx = 33
or, 331
313
3
xx
xx
or, 333313
3 x
x 33 1
xx = 0
or, 3
6 1
x
x = 0 x6+ 1 = 0 x6= 1
Now,x72+ x66+ x54+ x36+ x24+ x6+ 1
= (x6)12+ (x6)11+ (x6)9+ (x6)6+ (x6)4+ x6+ 1
= (1)12+ (1)11+ (1)9+ (1)6+ (1)4+ (1) + 1
= 1 1 1 + 1 + 1 1 + 1
= 1
52. (A) Let x = 3/13/1 )75627()75627(
x3 = 7562775627 +
1/3 1/33(27 756) (27 756)
)}75627()75627{(
x3 = 54 + 3(729 756)1/3x
= 54 + 3 (27) x
x3 = 54 9x
x3+ 9x 54 = 0
(x 3)(x2+ 3x+ 18)
x= 3
53. (D) a2d2+ b2c2 2abcd+ a2c2+ b2d2+ 2abcd
= a2(c2+ d2) + b2(c2+ d2)
= (a2+ b2)(c2+ d2)
= 2 1 = 2
54. (B) 347833
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= 2)32(833
= )32(833
= )38163(3
= 2)34(3
= 343
24
55. (A) x = 321
x 1 = 3 2 1
1
x 3 2
So,1
1
xx = 1 2 3 3 2
= 132
56. (B) n = 347
= 3434
= 322)3()2( 22
= 2)32(
So,
nn
1
= 32
132
=32
1347
=)32(
)32(4
= 4
57. (A)1
2
3
1
)2(
)1(
K
K
or, 3(K 1) = 2 K
or, 3K + 3 = 2 K
or, 3K + K = 2 3 2K = 1
K =2
1
58. (B) a+ b+ c= 6
On squaring both sides, we get
(a + b + c)2= 62
or, a2+ b2+ c2+ 2(ab+ bc+ ca) = 36
or, 14 + 2(ab+ bc+ ca) = 36
ab+ bc+ ca=36 14
2
i.e. ab+ bc+ ca= 11
Now,
a3+ b3+ c2 3abc = (a + b + c){a2+ b2+ c2
(ab + bc+ ca)}
i.e. 36 3abc= 6(14 11)
3abc= 36 18
abc= 6
59. (D) (x2+ 5x+ 10)-1=)105(
12 xx
)105(
12 xx
will be maximum when
x2+ 5x+ 10 is minimum and maximum
value of x2+ 5x+ 10 =
4
101452
=4
15
So, Required maximum value =15
4
415
1
60. (D) 2222 cosseccossin ec
22 cottan
= )tan(sec)cos(sin 2222
2222 cot)cot(costan ec
22 cottan
= )cot(tan2111 22
= }cottan2)cot{(tan23 2
= }2)cot{(tan23 2
= 2)cot(tan243
= 7)cot(tan27 2
0)cot(tan2
61. (D)x
xP
sin1
sin1
P=
x
x
cos
sin1
andx
xQ
cos
sin1
x
x
x
xR
sin1
sin1
sin1
cos
=x
xx2cos
)sin1(cos
R= 1 sinx
P= Q= R
62. (*)
tan57 cot37 cot33 tan53
tan33 cot53 tan33 cot53
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=
53tan
)155tan33(tan33tan
)33tan53tan1(
53tan
133tan
5333tan
1
= 33tan
53tan= 33cot53tan
= 57tan37cot
63. (D) 2 3sin sin sin 1
3 2sin sin cos
2 2cossin (1 sin
22 cos)cos2(sin
222 coscos2(cos1
4222cos]cos4cos4)[cos1(
6224 coscos4cos4cos4
44 coscos
04cos8cos4cos 246
4cos8cos4cos 246
64. (B) Let x = sin cos
x2 = cossin2cossin 22
x2 = 1 2sin cos
Now, 0 00 90
sin 1 and cos 1
sin cos 1
0 2sin cos 2
1 0 1 2sin cos 1 2
31 2 x
1 3x
1 sin cos 3
sin cos 1
65. (B) 0 0 0 0sin10 sin30 sin50 sin70
0 0 0 0 0 0 0sin(90 80 )sin(90 40 )sin(90 20 )sin30
1cos20 cos40 cos80
2
We know,
1
cos cos2 cos4 cos34
1
cos20cos40cos802
01 1cos(3 20 )
4 2
1
cos60
8
1 1 1
8 2 16
66. (B) 2sec 1 3 tan 3 1 0
21 tan tan 3 tan 3 1 0
2tan 3 tan tan 3 0
tan tan 3 1 tan 3 0
tan 3 tan 1 0
tan 3 0
tan 3
67. (C)
60
45
B
D
C
A
xmtr.
5000mtr.
060ACB
045DCB AB = 5000 mtr.
Let, AD= xmtr. From ABC ,
0
tan60AB
BC
5000
3BC
mtr. BC=5000
3mtr.
From ,DBC
0 5000
tan453
DBDB BC
BC
Now, AD = AB - BD
5000
5000 3
1
5000 13
mtr.
68. (D)
90 A A
2 2
A
2hh
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MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR
DILSHAD GARDEN ROHINI
BADARPUR BORDER
2tan
2
h h
A A ...(i)
2
tan 90
2
h
A
4cos
h
A
tan4
A
h ...(ii)
From (i) & (ii)
2
4
h A
A h
8h2= A2
h2= A2
8
22 Ah
2 2
Ah
69. (D)2
2
12sin
2
xx
x
or, 21
2sin2
2
xx
x
a b a b ab 2 2 2
[ ( ) 2 ]
1
2sin2
12
x
xx
021
xx
1
2sin
= 0 Also,
2sin1
x
70. (D)DE
AB
DEFPer
ABCPer
)(
)(
5.6
1.9
25
)(
ABCPer
Per (ABC) = 35
71. (B)
B
D
A
C
ABCADB
AC
AB
AB
AD
AB2= AD AC
72. (B) Let side of D = x
3243)2(
43 22 xx
32444
3 22 xxx
4x +4 =8
4x =4
x =1
73. (B)
B C
A
In ,ABC
AB + BC = 12 cm
BC + CA = 14 cm
& CA + AB = 18cm
2(AB + BC + CA) = 44 cm
AB + BC + CA = 22 cm
r
Now,
ATQ
22cm2 r
cm227
222
r7
2 cm
74. (D) Sum of all interior angles
= 2 sum of all exterior angles
36021802n
or, 7201802n (n 2) = 4
n = 6
Required no. of sides of the polygon = 6
75. (B)
In an equilateral triangle,
side = 32 inradius
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MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR
DILSHAD GARDEN ROHINI
BADARPUR BORDER
i. e. a= r32
= cm332
= cm36
76. (D)
B
A
D
5 cm7 cm
C
Here,
AD = angle bisector of A
DACBAD
In such case,
7:5
7
5
AC
AB
CD
BD
77. (*) No. of diagonals of a polygon of nsides
= nnC 2
ATQ,
nnC 2 = 54
n
n2 2 = 54 + 1
n n n
n
( 1) 2
(2 1) 2
= 54 + n
n2 n = 108 + 2n
n2 3n 108 = 0 n= 12, 9
n= 12
[ncannot be negative.]
78. (B) A35
B
P
Here, in ,AOP
AO = OP
35APOPAO
)352(180AOP
= 110
70110180POB
Also, In POB ,
BO = OP
552
70180OPBPBO
55ABP
79. (B) A
B C
I
65ABC
5.322
ABCIBC
Also, 55ACB
5.272
ACB
ICB
)ICBIBC(180IBC
= 180 60
= 12080. (C) Ratio of no. of sides = 1: 2
Then, Let the nos. of sides are nand 2n
Now, Ratio of their interior angles = 2 : 3
n
nn
n
2 490 2
4 4 390
2
n
n
2 4 1
4 4 3
6n 12 = 4n 4 2n= 8 n= 4
Respective nos. of sides of these polygons are 4 and 8.
81. (B)
:66
8
a
3
4
a
2
a
3
4
a3
3
3
3
82. (?) 4 r2= 2 (r + h2) = r(l+ r)
or, 4r = 2(r + h2) = l+ r
Now, 4r = 2(r + h2)
or, 4r = 2r + 2h2
or, 2r = 2h2
r : h2= 1 : 1
Again, 4r = l+ r
or, 4r = 232 hr + r
or, 3r = 232 hr
9r2 = 232 hr
8r2 = 23h
r22 = h3
r: h3= 1 : 22
r(h1) : h
2: h
3= 1 : 1 : 22
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MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR
DILSHAD GARDEN ROHINI
BADARPUR BORDER
83. (C)
5x
12x
13x
Let the sides are 13x, 12xand 5x.
ATQ,
13x+ 12x+ 5x= 450
or, 30x = 450 x= 15
Sides are of length 195m, 180m and 75m.
Now,
Ratio of sides are 13 : 12 : 5
(Pythagorian triplet)
Triangle is a right angled triangle.
So,
Area of triangle =2
1 base height
=2
1 12x 5x
=2
1 180 75 m2
= 6750 sq. meter
84. (A)
r = 10 cm
h = 4 cm
Let radius is increased by xcm.
New volume of cylinder = 4)10( 2 x
Again,
Let the height is increased by xcm.
New volume of cylinder = )4(102 x
4)10( 2 x = )4(102 x
(10 + x)2 4 = 100(4 + x)
(10 + x)2 = 25(4 + x)
100 + x2+ 20x= 100 + 25x
x2 5x = 0
x(x5) = 0
x = 5 cm
85. (B)
10 cm
10 cm
44 cm r
Volume of the
cylinder = hr2 r2 = 44 cm
= 1072 r= 222
744
cm
= 10497
22 r= 7 cm
= 1540 m3.
86. (C) Volume of cube = a3(where a = length ofa edge)
When each edge is increased by 40%.
length of the new edge = 1.4a Volume of new cube = (1.4a)3
= 2.744 a3
Required % increase
= 3
33744.2
a
aa 100%
= (1.744 100)%
= 174.4%
87. (B)
h = 24cm
r
l
22 rhl +=22 724 +=
= 25 cm
Volume of cone = 1232 cm3
hr2
3
1 = 1232 cm3
247
22
3
1 2 r = 1232 cm3
r2 =2422
371232
r2 = 49 cm
r = 7 cm
So, Curved surface area of cone = rl
= 2577
22 cm2
= 550 cm2
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DILSHAD GARDEN ROHINI
BADARPUR BORDER
88. (D)
18 cmr
26 cm
Perimeter of the rectangle= 2 (26 + 18) cm
= 88 cm
= r2 r =222
788
= 14 cm
Area of the circle i.e. 2r
=7
22 14 14
= 616 cm2
89. (A) A
B C
a
16 cm 25 cm
28 cmO
a
a
Area of ABC
= Area of )( BOCAOCAOB
2
4
3a = )282516(
2
1a
a= 6923
4
cm
a= 346 cm
So,
Area of ABC =2)346(
4
3
= 321164
3
= 31587 cm3
90. (C)emptyspace
h
Volume of water needed to fill the emptyspace
= Volume of cylinder Volume of cone
= hrhr22
3
1
= hr2
3
2
=
hr2
3
12
= 272 cm3
= 54 cm3
91. (B)
7cm
7 cm
7cmAC
In right angled isosceles ABC,
AB = BC
Also,
AD = CD = BD = 7cm
= circum radius
(where ACBD )
Now, Required area= Area of semicircle Area of ABC
=
714
2
1
27
7722cm2
= (77 49) cm2
= 28 cm2
92. (B) A
B CD
EF
G8
53
6 4
10
In the given ABC ,
AD, BE and CF are medians and they cut
one another at G.
GF
CG
GE
BG
GD
AG =
1
2
Here,
AD = 9 cm, BE = 12 cm and CF = 15 cm
AG + GD = AD = 9 cm
AG = 6 cm and GD = 3 cm
Also,BG + GE = BE = 12 cm
BG = 8 cm and GE = 4 cm
Also,
CG + GF = CF = 15 cm
CG = 10 cm and GF = 5 cm
Area of AGB = 862
1 = 24 cm2
So, Area of ABC = AGB3
= 2 24 cm2
= 72 cm2
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DILSHAD GARDEN ROHINI
BADARPUR BORDER
93. (C) T.S.A. of prism = C.S.A. + 2 Area of
base
608 = Perimeter of base height + 2
Area of base
608 = 4x 15 + 2x2
(where x= side of square)
x3+ 30x 304 = 0
(x 8) (x+ 38) = 0
x= 8
Volume of prism = Area of base height
= 8 8 15
= 960 cm3
94. (A) 22)1( rr = 22
)12( 22 rrr
r2
= 22
7
22(2r+ 1) = 22
2r+ 1 = 7
2r= 6
r= 3 cm
95. (B) Volume of water due to 2 cm rain on a
square km land
= 1km 1 km 2 cm
= m100
2m1000m1000
= 20000 m3
50% of volume of rain drops
= 10000m3
Now,
Required level by which the water level
in the pool will be increased
=m10m100
m10000 3
= 10m
96. (A) Total income = Rs. 150000
Required difference= (15% 5%) of Rs. 150000
= Rs. 15000
97. (C) Maximum expenditure of the family
other than on food, was on others.
98. (B) Saving = 15% = Housing
99. (D) Required % = 10% + 12% + 5%
= 27%
100. (D) Money spent on food
= 23% of Rs. 150000
= Rs. 34500
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SSC (Tier-II) - 2013 (Mock Test Paper - 2) (ANSWER SHEET)
1. (C)2. (D)3. (D)4. (C)5. (A)6. (B)7. (C)8. (C)9. (D)10. (B)11. (C)12. (*)
13. (A)14. (B)15. (C)16. (C)17. (A)18. (D)19. (C)20. (A)
21. (B)22. (B)23. (D)24. (*)25. (C)26. (A)27. (B)28. (B)29. (D)30. (D)31. (A)32. (C)
33. (A)34. (D)35. (D)36. (B)37. (*)38. (B)39. (C)40. (B)
41. (C)42. (C)43. (B)44. (D)45. (A)46. (*)47. (B)48. (A)49. (C)50. (D)51. (A)52. (A)
53. (D)54. (B)55. (A)56. (B)57. (A)58. (B)59. (D)60. (D)
61. (D)62. (*)63. (D)64. (B)65. (B)66. (B)67. (C)68. (D)69. (D)70. (D)71. (B)72. (A)
73. (B)74. (D)75. (B)76. (D)77. (*)78. (B)79. (B)80. (C)
81. (B)82. (*)83. (C)84. (A)85. (B)86. (C)87. (B)88. (D)89. (A)90. (C)91. (B)92. (B)
93. (C)94. (A)95. (B)96. (A)97. (C)98. (B)99. (D)100. (D)